Re: help with a basic differentiability and derivative function

David L. Wilson wrote:
I am replying to my own post to add an additional final paragraph to further clarify.

"David L. Wilson" <dwilson314@xxxxxxxxxxxx> wrote in message news:Y6Gdna_kruMIWTXenZ2dnUVZ_sOdnZ2d@xxxxxxxxxxxxxxx

"David Wilkinson" <david@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message news:do92n6$p1r$1@xxxxxxxxxxxxxxxxxxxxxxx

David L. Wilson wrote:

"David Wilkinson" <david@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message news:do8f40$pjb$1@xxxxxxxxxxxxxxxxxxxxxxx


disanalysis wrote:


very basic problem...can anyone help me with the proof?

f(x) = x^2-3x+8 if x>1
f(x) = 2x^2+x+3 if x=<1

using epsilon-delta property can you prove that f(x) is differentiable
at all points except x=1
and that it is not differentiable at 1?

thanx


At x = 1 f(x) = 6 from both expressions.

for x > 1, df(x)/dx = 2x-3 -> -1 as x -> 1 from the first expression, and

For x <= 1, df(x)/dx = 4x+1 = 5 at x=1.

So the first derivative of f(x) is discontinuous at just above x = 1. But, at x = 1 it is differentiable from the second expression which holds up to and including x = 1.


Wrong. The derivative does NOT exit at x=1 because the limit definining the derivative at that point does not exist as the limit defining the derivative at x=1 from the left is not equal to the limit defining the derivative at x=1 from the right. Or were you intentionally giving him a wrong answer on this obvious homework problem.

Graphically, there is a sharp turn at x=1, hence, the derivative does not exist at the point (1, 6) although the function is continuous there.

Right. Since the second expression holds up to x = 1 the derivative is defined by it at x = 1. The first expression only holds for x > 1 so the derivative is different for x > 1, not at x = 1.

NO NO NO
The derivative at a point is defined as a limit (the usual definition in a calculus text). That limit exists if and only if that limit from the right and that limit from the left are the same. In this case, that limit from the left is 5 and that limit from the right is -1 so the limit that defines the derivative at x=1 does not exist. However, one can say at x=1, the left derivative is 5 and the right derivative is -1.

The fact that one of his equations defines the function for x<=1 does not allow one to use it alone to define the derivative at x=1----Your sentence "Since the second expression holds up to x = 1 the derivative is defined by it at x = 1." is terribly wrong.


To look at it another way...Change the definition of the function to
f(x) = x^2-3x+8 if x=>1 (I moved the equal to here.)
f(x) = 2x^2+x+3 if x<1
This is the same function. Each x has the same f(x) value. But by your argument the derivative at x=1 is now -1 instead of 5---but it is the same function. Something must be wrong (namely the derivative at x=1 actually does not exist).


No, that is not the same function. You have changed the value of f(x) at x = 1 from 2x^2+x+3 to x^2-3x+8 so it is a different function with a different slope at x = 1.

As I understand it, if a function g(x) has continuous ordinate and first derivative at x then one can define the first derivative as any of

Lim delx -> 0 (g(x+delx)-g(x))/((x+delx)-x) or

Lim delx -> 0 (g(x)-g(x-delx))/(x-(x-delx)) or

Lim delx -> 0 (g(x+delx)-g(x-delx))/((x+delx)-(x-delx))

and they would all give the same answer.

However, in this case the first derivative of f(x) is discontinuous at just above x = 1, not at x = 1, so the first expression for the derivative is appropriate for x > 1 and the second for x <= 1. The third expression is not useful in this case because of the discontinuity at just above x = 1.
.